SPECIAL POSITIONS FOR ESSENTIAL TORI IN LINK …jb/essential-tori.pdf · SPECIAL POSITIONS FOR ESSENTIAL TORI IN LINK COMPONENTS 527 Example of type 1 embedding Fig. 2. Example of

Pergamon Topology Vol. 33, No 3, pp. 525 ~556, 1994

Copyright 8 1994 Elsevier Science Ltd

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SPECIAL POSITIONS FOR ESSENTIAL TORI IN LINK COMPLEMENTS

JOAN S. BraMAN? and WILLIAM W. MENASCO~

(Received 15 September 1991; in revised form 17 April 1993)

INTRODUCTION

THE decomposition of links into non-split components, by cutting along essential 2-spheres,

is a fundamental step in any attempt to understand the link problem. In an earlier paper [2]

the authors studied that problem from the point of view of braid theory. A somewhat more

subtle decomposition of a link complement involves splitting along essential tori. By a

fundamental theorem which is due to Alexander (see p. 107 of [S]) every embedded torus T

in S3 is the boundary of a solid torus V on at least one side. The solid torus V may, however,

be knotted, and this makes the study of embedded tori much more difficult than embedded

2-spheres, since the latter cannot be knotted. The first major attempt to understand

embedded tori in link complements was a groundbreaking paper by Schubert [6]. The

seminal role which is played in the topology and geometry of link complements by

embedded tori was later underscored in the important work of Jaco and Shalen [3],

Johansson [4] and Thurston [7], who showed that if M3 is a 3-manifold, then there is a

finite collection R of essential, non-peripheral tori T,,. . .,T, in M3 such that each

component of M3 split open along the tori in 0 is either Seifert-fibered or hyperbolic. Our

goal in this paper is to apply the techniques of [2] to the study of essential tori in link

complements.

Let II be a link type in S3, with representative L. A torus T in S3-L is essential if it is

incompressible, and peripheral if it is parallel to the boundary of a tubular neighborhood of

L. A link type E. is simple if every essential torus in its complement is peripheral, otherwise

non-simple. Satellite links (defined below) are a special case of non-simple links.

To describe our results, assume that L is a closed n-braid representative of 1, with braid

axis A. The axis A is unknotted, so S3-A is fibered by open discs {H,; 8 E [0,2rc]}. It will be

convenient to think of A as the Z-axis in R3, and the fibers H, as half-planes at polar angle 8.

Then A and the half-planes H, serve as a “coordinate system” in R3 which can be used to

describe both L and T. We call our canonical embeddings types 0, 1 and k, where in the

latter case k is an integer r 2. Type 0 will be familiar to most readers, but types 1 and k do

not appear to have been noticed before this as general phenomena:

Type 0. The torus T is the boundary of a (possibly knotted) solid torus V in S3, whose

core is a closed braid with axis A. The link L is also a closed braid with respect to A, part of

it being inside V and part of it (possibly empty) outside. The torus T is transverse to every

fiber H, in the fibration of S3-A, and intersects each fiber in a meridian of V. It is foliated by

tPartially supported by NSF Grant DMS-92-06584 and by the US-Israel Binational Foundation Grant 89- 000302-2.

fpartially supported by NSF grant DMS-92-00881.

525

526 Joan S. Birman and William W. Menasco

Example of type 0 embedding

Fig. 1

these meridian circles. An example is given in Fig. 1. In this example (and in all other braid

diagrams in this paper) the braid strands are to be regarded as being weighted by arbitrary

positive integer weights wi, where weight wi means “replace the single strand by wi parallel

strands”. Notice that the axis A does not intersect V.

Type 1. Choose a a-sphere S which is pierced twice by A and has a standard north-

south foliation. Choose two points of S\S n A, and join them by an arc LX, chosen so that

interior of a intersects fibers of H transversally. The torus obtained by tubing S to itself

along a neighborhood of cz, is a type 1 torus. An example is given in Fig. 2. The axis pierces T twice in the “sphere part” of T. In general the “tube part” of T could go around A several

times as a braided tube. The axis A intersects the obvious solid torus V in a single arc.

Type k (k 2 2). The torus T is made up of k 2 2 cylinders which are glued together in a

cycle. The core of the ith cylinder is an arc tli which lies entirely in a fiber of H and has its

endpoints on A. Thus each cylinder intersects A twice, and JT n Al = 2k. An example (for

k = 2) is given in Fig. 3. In this example T bounds a visible unknotted solid torus V, but in

general V will be knotted. (Peek ahead to Fig. 8 for an example where the arcs CI~,. . .,CQ

which are the cores of the cylinders fit together to give the trefoil knot.) In every case A n V

will be a union of k unknotted, unlinked, disjoint arcs.

If we begin with an arbitrary closed braid representative L, then the tori T,,. . .,T, which occur in the Jaco-Shalen-Johansson decomposition will in general not be isotopic in

the complement of L to tori of types 0,l and k. However, if we modify L in an appropriate

way, it may be possible to isotope them into the standard position. We will prove that the

modifications which are needed are a finite sequence of “exchange moves”, where an

exchange moue is defined pictorially by Fig. 4. Notice that exchange moves take closed n-

braid representatives of 2 to other closed n-braids. To set the stage for our work here, we

note that in [2] we showed that every closed n-braid representative of a prime or split link

was exchange-equivalent to a closed n-braid representative which is “obviously split or

composite”. Our principal result is that, up to exchanges, every torus in the

Jaco-Shalen-Johansson decomposition may be assumed to be embedded in one of our

three canonical ways:

THEOREM 1. Let R = {T, ,. .,T,} be the finite collection of tori in the Jaco-Shalen-Johansson decomposition of S3-L. Then there is a finite sequence of n-braids:

SPECIAL POSITIONS FOR ESSENTIAL TORI IN LINK COMPONENTS 527

Example of type 1 embedding

Fig. 2.

Example of a type k embeddlng (k=2)

Fig. 3.

L = L, -+ L, -+ . . . . . + L, = L*

such that each Li+ I is obtained from Li by an exchange move together with an isotopy in the

complement of A, such that in the braidjibration associated with L* each torus in the modified

collection R* is isotopic in the complement of L* u A to a torus of type 0, 1 or k (k 2 2).

Implicit in Theorem 1 is a description of all non-simple links, via their closed braid

representatives. We are able to make this explicit in the special cases of links of braid index 3

and 4. Let oi denote an elementary braid in which the ith braid strand is interchanged with

the i + 1st.


The Exchange

Fig. 4.

COROLLARY 1. Let L be a closed 3-braid representative or a prime link of braid index 3,

and let T be a nonperipheral essential torus in S3-L. Then T has a type 0 embedding, and L is

conjugate to:

(a,)p(a,a:a,)q> where IpI 2 2, 141 2 1.

Let oij, denote an elementary braid in which the ith and jth strands exchange positions,

the jth crossing over the ith, and both passing in front of all intermediate braid strands as

they do so.

COROLLARY 2. Let L be a 4braid representative of a prime link of braid index 4, and let T

be a nonperipheral essential torus in S3-L. Then after a series of exchange moves, the

conjugacy class of L and type of T may be assumed to be one of the following,

Type 0. (cr,)P(a3)q(o,o,030,)r, where IpI > 0, (41 > 0, and IrI 2 2, or

lj(~~~~~~~~cr~)~, where IpI 2 1, for any 3-braid fi = W(a,, a,), or

(~,)p(~,~:~,)q(~3~:~3)r, where IPI, 141, Id > 0.

Type 1. ~(o,a~a,)P, IpI 2 1, for any 3-braid /I = W(o,,cr,),

Type 2. (~13)p(~24)r(~13)q(~24)s, where IPI, Id I4 Isl >O.

Our second set of results concerns numerical data which appears as a result of the

partial normal forms of Theorem 1. Let L be a link which is represented as a closed braid,

relative to a choice of braid axis A and fibration H. Suppose that there is an essential, non-

peripheral torus T in S3-L. By a well-known theorem of Alexander, T bounds a solid torus

V on at least one side. In general, the braid axis A will intersect V in a finite family of disjoint

arcs, and some features of the closed braid representation of L can be deduced by

investigating the relative local positions of L n V and AnV in V.

THEOREM 2.

(i) Assume that T has a type 0, 1 or k foliation. Let V be the solid torus which T bounds.

Then A A V is a union of 0, 1 or k 2 2 arcs, according as T has a type 0,l or k foliation, where

in the situation of type 0, z~T bounds on both sides we choose V so that A n V is empty.

(ii) The inclusion of (L u A) n V in V in the 3 cases is as depicted in Fig. 5. Each

component of A n V is illustrated in the projection as a heavy black dot, and L n V wraps


around these arcs to giae “local” braid diagrams. The weights w,,, w1 ,. . .,wk are the number of

braid strands qf L in the local pictures in Fig. 5 (also see the examples in Figs 1, 2, 3).

(iii) If it should happen that V is unknotted, so that V’ = S3 - V is also a solid torus, then

a symmetric description holds for the inclusion of (L u A) n V’ in V’, for types 1 and k.

As an application of Theorem 2 we consider the special case of satellite links. Let V,,, be a

standard model for an embedded solid torus in R3, say a torus of revolution. Let M be a

non-trivial knot or link which is embedded in a geometrically essential manner in V,,,, i.e. M

meets every meridian disc of V,,, non-trivially, also M is not isotopic to the core of V,,,. Let C

be any non-trivial knot in R3, and let V, be a solid torus neighborhood of C. Let h: V,,, -+ V,

be a homeomorphism, and let L = h(M). Then C is a companion of the satellite link L, and

M is the model for L. Notice that satellite knots always have an incompressible torus in their

complement, i.e. r?V,. The fact that, by hypothesis, M is not isotopic to the core of V,,, ensures

that T, = 3’, is not peripheral. The fact that, by hypothesis, C is knotted ensures that T,

does not bound a solid torus on both sides.

If L represents a link type, its braid index b(L) is the smallest integer n such that L is

isotopic to an n-braid. Our application is a new proof and generalization of a theorem of

Schubert [6] about the braid index of satellite links. Recall that in Fig. 5 and also in the

Type 0 Type 1

Type k with kz4

A)IIV for the three types of embeddings

Fig. 5.

of T=aV

530 Joan S. Bit-man and William W. Menasco

examples in Figs 1, 2, 3 we defined the weights:

w0 in the case where T, has a type 0 embedding,

wO, w1 in the case where T, has a type 1 embedding, and

w1 ,. . .,wk in the case where T, has a type k embedding.

COROLLARY 3. t Let L be u sutellite knot with model M and companion C. Let T, = N,.

Then:

b(L) = w,.b(C) if the embedding of T, is type 0. (cf Satz 23.2 of [6])

b(L) = w,.b(C) + w1 if the embedding of T, is type 1.

b(L) = w1 + w2 +. . . + wk if the embedding of T, is type k 2 2.

The proof of Corollary 3 will follow directly from the machinery we develop to prove

Theorems 1 and 2. We sketch it now, because it shows the usefulness of the structure

provided by our “coordinate system”, i.e. the braid axis and the fibers of H.

Proof of Corollary 3. Let b(L) be the braid index of L. We choose a closed b(L)-braid

structure for L, with axis A. The complement of the axis A is an open solid torus, which is

fibered by meridian discs {H,; 8 E [0,2n]}. The link L intersects each fiber H, of H in b(L)

points, which we refer to as “dots”. Our task is to count the dots. By Theorem 1 we may

assume that T, has an embedding of type 0, 1 or k relative to the fibers {H,; 8 E [0,271]} in

our braid structure. This means that if H, is non-singular, then T, n H, will be a family of

arcs and circles. The idea is to choose H, so that the arcs and circles subdivide the dots into

groups which are convenient for counting purposes. See Fig. 6.

Type 0. There are no singular fibers, and any H, will do. The fact that the braid index of

L is minimal implies that the core of V, must be a closed b(C)-braid with respect to A. The

intersections of T, with H, will therefore be a collection of b(C) circles. Each circle bounds a

disc d, in H,, and d, contains w0 dots. Thus b(L) = w,.b(C), as claimed.

Type 1. The incompressible torus which is being studied has a natural decomposition as

a union of a single punctured 2-sphere S with a braided tube attached to it. The sphere and

tube are joined along an annulus A. We choose our fiber H, so that it does not meet the

annulus. With this choice T, n H, will consist of a single arc and n circles c1 ,. . .,c,, if C is

represented as an n-braid. The arc bounds a half-disc which contains w. dots. The total

number of dots will be (w,.n + MI,). The integers \vO and w, are both invariants of the model

M, thus the braid index of the satellite will be minimized when the braid index of the

companion is minimized. Thus b(L) = wO. b(C) + wl, as claimed.

Type 0 Type 1

Intersections of T

special non-singular

Fig. 6.

See the note, added in proof. at the end of the paper.

Type k (k=4)

and L with a

fiber H, of H


Type k. The incompressible torus which is being studied has a natural decomposition as

a union of k twice-punctured spheres S,, . . , S, joined up in a cycle by annuli A 1, . . . ,A,.

We choose H, so that it does not intersect the annuli. With this choice T, n H, will consist of

exactly k arcs, b,, . . . ,bk, where the ith arc cuts off a half-disc which does not contain any

other bj. (Every non-singular fiber intersects T, in a family of k arcs, but in general both half-

discs determined by a typical arc bi will contain other arcs bj). The half-disc contains wi dots.

Thus b(L) = w1 + w2 + . . + wk, as claimed. II

A doubled knot L is a satellite knot which is based upon the model which is illustrated in

the top right sketch in Fig. 7. Its companion is an arbitrary non-trivial knot. In the example

in Fig. 7 it is a trefoil. Our final result concerns the essential torus T, in the case when L is a

doubled knot.

COROLLARY 4. Let L be a doubled knot with companion C. Assume that L is represented in any way as a closed braid. Then after a series of exchange moves the essential torus T, = aV, in S3 - L will have a type k embedding.

Proof of Corollary 4. Let V, be a tubular neighborhood of the companion, so that V,

contains L. Let T, = 8V,. Notice (consult the model) that L intersects each meridian disc of

Fig. 7.


V, zero times algebraically. Choose a closed braid structure with axis A and fibration H for

L. By Theorem 1, we may assume without loss of generality that T, has an embedding of

type 0,l or k relative to H. However, from Fig. 5 we conclude that T, cannot be type 0 or 1,

because in each of those cases there is a meridian disc of V, which intersects L with non-zero

algebraic intersection number. Thus T, must have a type k embedding. II

Examples of Corollary 4 are interesting and subtle. Figure 7 gives one, when L is a

double of the trefoil knot C. In this example k = 5. This example illustrates that the formula

in Corollary 3 for b(L) in the case of type k embeddings is more complicated than it appears

to be, because in Fig. 5 we are looking at the inclusions of (L u A) n V, in V,, and this

contains more structure than L n V,, which is homeomorphic to M n V,. For example, in

the situation of the doubled knot which is illustrated in Fig. 7 there is a model M which

contains 2 blocks. However, when we add A we see that we need 5 braid blocks, not 2. The

additional blocks correspond to additional braid crossings which cancel as braid crossings,

but which are needed in order to give the correct embedding of L u A in V,. The structure which is imposed by doubling a knot is so very far away from the structure

imposed by a braid axis, that one would not expect that anything at all could be said about

the braid representatives of doubled knots. However, the work we have done suggests a

conjecture:

Conjecture: Let D(C) be a doubled knot with companion C. By Corollary 4 the natural

torus T, = iW, in S3 - D(C) has a type k embedding. The example in Fig. 7 then suggests

that when D(C) is represented as a closed braid with a minimum number of braid strands,

its companion C will have a projection which exhibits it as a union of k unknotted and

unlinked arcs c(r) . . ,Q, each of which has its endpoints on A. See Fig. 8 for an example, in

the case of the trefoil. The smallest integer k such that C can be so-represented is (by

definition) a numerical knot invariant k(C) of the knot type of C. It is reasonable to

conjecture that the braid index of D(C) is determined by the integer k(C) and the framing.

The invariant k(C) seems not to have been noticed.

Fig. 8.


The organization of this paper is as follows. In $1 we introduce our basic machinery,

which will involve the study of certain foliations of essential tori. Tori which admit a circular

foliation will be seen to have an embedding of type 0, but the other cases are more

complicated. In $2 we standardize the foliations for tori which have tilings, and show that

after this standardization the embedding is type k. In $3 we standardize the mixed foliations,

and show that (after further modifications) they have type 1 embeddings. In 94 we use all of

this to prove Theorem 1, Corollaries 1 and 2 and Theorem 2.

$1. FOLIATIONS ON TORI

The techniques used in this paper are related to techniques used in [2]. In this section we

give a quick review of the relevant material, with references, as we need them, in lieu of

repeating details of arguments which are published elsewhere.

Our work begins with a non-simple link type 2, a closed n-braid representative L, and an

essential non-peripheral torus T in S3 - L. The closed braid structure is defined relative to a

braid axis A and a choice H of fibration of S3 - A by meridian discs {H,; 6~ [0,27c]}. The

intersection of the H,‘s with T induce a foliation of T. In this section we begin to standardize

the foliations.

LEMMA 1. We may assume that:

(i) The intersections of A with T arejnite in number and transverse. Also, ifp~A n T,

then p has a neighborhood on T which is radially foliated by its arcs of intersection withfibers of H.

(ii) All butfinitely many3bers H, of H meet T transversally, and those which do not (the singular Jibers) are each tangent to T at exactly one point in the interior of both T and H,. Moreover, the tangencies are local maxima or minima or saddle points.

Proof: Use general position. II

A singular leaf in the foliation of T will be one which contains a point of tangency. All other leaves are non-singular. It follows from (ii) that:

(iii) Each non-singular leaf is either an arc or a simple closed curve. (iv) A singular fiber contains exactly one singular point. (v) Each singular point is either a center or a saddle.

We refer to the leaves of the foliation of T as b-arcs and c-circles. Each b-arc and each c-

circle lies in both T and in some fiber He of H. A b-arc b is essential if both sides of H, split

along b are pierced by L. A c-circle, c in T A HO, is essential if c bounds a subdisc d, of H,

such that d, A L is nonempty.

LEMMA 2. Assume that T satisfies (i)-(v) and has b-arcs or c-circles in the induced foliation. Then we may replace T with an essential torus, T’, such that the foliation of T’ also satis$es (i)-(v) and:

1. All b-arcs are essential.

2. All c-circles are essential. 3. Any c-circle in the foliation is homotopically non-trivial on T.

Moreover, if L is not a split link then T’ is isotopic to T.


ProcJf

1. Suppose that h is an inessential b-arc in a fiber H, of H. Then one of the sides of H,

split along h is a disc d, which is not pierced by L. Without loss of generality we may assume

that h is innermost. We may then push T across A in a regular neighborhood of d, in 3-space

to remove the b-arc from H, and all nearby fibers.

2. To show that we can remove inessential c-circles we reason as Bennequin did, in [l].

3. Suppose that c is an essential c-circle in a fiber H, of H. We may assume without loss

of generality that c is inner-most in H,. Then c bounds a sub-disc d, of H,, which is

punctured by L algebraically non-zero times. If c is homotopically trivial on T then c

bounds a sub-disc of T, d,. We can construct a 2-sphere by gluing d, to d, along c. But this

will be punctured algebraically non-zero times of L, which is impossible for a 2-sphere in

3-space. Thus, c must be homotopically non-trivial on T. II

We now consider the different types of singularities which can occur in the foliation.

Since there are only two types of leaves in the foliation there are at most three possible types

of singularities, which may be described as types bb, bc and cc, according as the leaves which

are surgered are both type b, types b and c, or both type c.

LEMMA 3. Singularities of type cc do not occur.

Proof: A surgery between two c-circles will produce a homotopically trivial simple

closed curve, violating assertion 3 of Lemma 2. II

We restrict our attention, temporarily, to the situation where every leaf in the foliation of

T is a b-arc. This is exactly the situation which was studied in [Z], for a foliated essential 2-

sphere X in a link complement. It was shown that in this situation there is a natural foliated

cell-decomposition of X. The O-cells of the decomposition are the points of A I-J X, and the 2-

cells are foliated squares, each of which contains exactly one bb-singularity. We call these

squares bb-tiles. The 1 -cells are any convenient choice of non-singular b-arcs in the foliation

of X which divide X into bb-tiles. See Fig. 9. The description of this cell decomposition goes

through in exactly the same way if we replace the 2-sphere X by our essential, non-

peripheral torus T. There is a natural foliated cell-decomposition of T, as before, and each 2-

cell has 4 sides and contains a single bb-singularity.

A related situation exists when both bb- and bc-singularities occur, only now we must

decompose T along a finite collection of non-singular b-arcs and c-circles into a union of bb-

tiles and bc-annuli, depicted in Fig. 9. Each bc-annulus contains two points of A n T in its

boundary, and exactly one singularity of the foliation in its interior. The singularity is type

bc. This leads us to the following definitions:

If the foliation of T consists entirely of c-circles, with no singularities, then T has a

circular ,fihution.

H@B bc-annulus

bb-tile

Fig. 9


If the leaves in the foliation of T include both b-arcs and c-circles, so that the

singularities include both bb- and bc-singularities, we say that T has a mixed foliation.

If every non-singular leaf in the foliation of T is a b-arc, and every singularity is a bb-

singularity, we say that T admits a tiling.

Now observe that if we have a collection 52 of finitely many tori instead ofjust one torus, all

of the modifications so far may be done on them, one at a time. Therefore, using the

notation just introduced, we may put together Lemmas l-3 to obtain:

LEMMA 4. We may assume that each torus Ti in R has either a circularfoliation or a mixed

foliation or a tiling. Moreover, if the.foliation is circulur, then Ti has a type 0 embedding.

Proof. Everything except the last assertion has been proved. So, assume that T, admits a

circular foliation. There are no singularities in the foliation of Ti and every leaf c0 is a circle.

Since each c, lies in a fiber H,, it bounds a disc d, in H,. These discs sweep out a solid torus

Vi and every leaf in the foliation of Ti is a meridian of Vi. The embedding is type 0. //

$2. TORI WHICH ADMIT TILINGS

Our goal in this section is to prove that if a torus has a tiling, we may modify it to one

which has a “standard” tiling. After the tiling is standardized we will be able to prove that

the embedding is type k, where k > 2.

Let T be a torus which admits a tiling. The valence of a vertex in the tiling of T is the

number of tiles which meet at that vertex. Assign a positive side to T, arbitrarily. The sign of

a tile is + (respectively minus) according as the sense of the outward pointing normal to T

at the singular point agrees (resp. disagrees) with the sense of normal to H, which points in

the direction of increasing 8. We say that a tiling is standurd if all of its vertices are valence 4,

and if the signs of the four tiles which intersect at a common vertex are + , - , + , - in that

cyclic order. (We will see later that this implies much more: There is a fundamental domain

for T which is a tiled rectangle of dimension 2 tiles by 2k tiles, with opposite sides identified.)

Our first goal in this section is to show that after a series of exchange moves the tilings of

Lemma 4 may be modified to standard tilings. Similar work was done in 121, when we

considered tilings of 2-spheres in the complement of a split link E. which is represented as a

closed braid. If X is a surface in 3-space which admits a tiling relative to H, we define the

complexity C(X, H) of the tiling to be the pair (V, R), where Vis the number of vertices and R

the number of tiles. We say that C(X’, H’) = C(X, H) if (V’, R’) < (V, R) using lexicographical

ordering. If we had realized when we wrote [2] that we would need Lemma 3 of that paper

for this paper, we would have stated it as follows:

LEMMA 5. (Lemma 3 of [2], restated). Let X be u surface which is in the complement ofL.

Assume that X has a tiling, with complexity C(X, H). Suppose there is a foliated subset N, of X

which is the star of a vertex p of valence v 2 3. Suppose that there are adjacent singularities s

and s’ in N, which have the same sign. Then there is an isotopy of X in the complement of L to a

surface X’, with C(X’,H) = C(X,H), such that the tiling of X’ is identical to that of X

everywhere except in N,, and in the new tiling the valence of p is reduced to v - 1. Moreover,

the isotopy is supported in a disc neighborhood N in N, of an arc p which joins s to s’in N,

chosen so that p crosses only non-singular leaves, and crosses each such leaf transversally.


Proof The assertion can be understood intuitively with the help of Fig. 10. The top

picture shows N, embedded in 3-space. Fibers of H are horizontal planes. The isotopy of X

may be visualized as a coalescing of the two singularities to a single monkey saddle, and

then a splitting apart in a new way. The bottom picture shows the changes in the foliation of

N, as the singularities coalesce and split apart again, in the case when N, is a union of two

adjacent bb-tiles.

See the proof of Lemma 3 of [2] for details. II

LEMMA 6. Suppose that T has a tiling which is induced by thejbration H of S3 - A. Then,

(i) If there is a vertex p of valence v in the tiling of T such that there are adjacent tiles

meeting at p which have the same sign, then there is an isotopy of T in the complement of L to

a tiled torus T’ such that C(T’,H) = C(T, H), and in the tiling of T’ the vertex p has

valence v - 1.

(ii) If there is a vertex of valence 2 or 3 in the tiling of T, the closed braid L admits an exchange. After the exchange there will be a new torus T’ with a tiling such that C(T’, H) < C(T,H).

Proof

(i) This is a consequence of Lemma 5 above.

(ii) First suppose that there is a vertex of valence 3 in the tiling of T. Either all of the tiles

have the sign, or two of them have one sign and the third the opposite sign. Thus we may

always find a pair of adjacent tiles with the same sign. By Lemma 5 above there is then an

-------I fc- 1

non-degenerate saddles at s and s' signs ats and s' agree

pjg$gJ+~-J@q+

+pjyj+m Changesin the foliation of R during

the passage from X to X'

Fig. 10.


isotopy of T to a new torus T’ such that the tiling of T’ has a vertex of valence 2, also

C(T’, H) = C(T ,H). We may thus assume that the tiling of T contains a vertex of valence 2.

This is the situation which was considered in Lemma 4 of [2]. The argument in Lemma 4 of

[2] related to the situation where there is a 2-sphere X (rather than an essential torus T) in

the complement of L, however it makes no use at all of the global topology of X, and the

arguments go over without any change at all when we replace X by T. By Lemma 4 of [2]

we conclude that L admits an exchange move, and also after the exchange move T can be

modified by isotopy to a new essential torus T’, such that C(T’,H) < C(T,H). II

LEMMA 7. Assume that T is a torus in the complement of a closed braid, and that T admits

a tiling. Then, after a sequence of exchanges we may assume that T admits a standard tiling.

Proof: As noted earlier, the tiling of T produces a natural foliated cellular decomposi-

tion of T. Let V, E and R be the number of 0,l and 2-cells. Then V - E + R = 0, because T

is a torus. Since each l-cell is the boundary of exactly two 2-cells, and each 2-cell has four l-

cells in its boundary, we have 2R = E. Thus 2V - E = 0. Now let V(i) be the number of

vertices having valence i. Since there are no vertices of valence 1, we have V = V(2) + V(3)

+1/(4)+.... Since each edge has 2 vertices in its boundary we have 2E = 2 V(2) + 3 V(3)

+ 4V(4) + . . . . Thus

2V(2)+ V(3)= V(5)+2V(6)+3I’(7)+.... (I)

By equation (l), if there are any vertices having valence other than four then there must exist

vertices of valence 2 or 3. It then follows from Lemma 6 that L admits an exchange move.

After the exchange, we may eliminate two points of A n T. Continue in this manner until

every vertex has valence 4.

Now suppose that there is a vertex p such that a pair of adjacent tiles in the cyclically

ordered array of 4 tiles about p which have the same signs. By assertion (i) of Lemma 6 we

may find a new fibration H’ such that the valence of p is reduced to 3. But then, after an

exchange we may reduce the complexity by eliminating two points of A n T. This process

ends with a torus T which has a standard tiling. II

LEMMA 8. A torus which has a standard tiling has a type k embedding.

Proof: We are given a torus T which admits a standard tiling, i.e. T is a union of bb-tiles,

every vertex in the tiling has valence 4, and the cyclically ordered array of signs of the four

tiles which meet at each vertex are + , - , + , - . We will show that T has a type k

embedding.

We begin our proof with an observation about the order of the vertices of a bb-tile

around the tile boundary and along the axis A. Let T be such a tile, and let 1, 2,3,4 be its

vertices, cyclically oriented around t3T, as in Fig. 11. The vertices represent points where the

braid axis A pierces the torus T, and we claim that the four vertices have the same cyclic

order 1, 2, 3, 4 (or its reverse) on A as on dT. To see this, suppose that the arrows on the

leaves of the foliation on T denote the sense of the flow as 0 is increasing. Choose two non-

singular leaves, say c1 and c(* which lie in the same fiber H, of H, before the singularity. Since

these leaves are disjoint arcs in H,, it follows that the endpoints 1 u 2 of u cannot separate

the endpoints 3 u 4 of cl* on A = aH,. So the cyclic order is either 1,2,3,4 or 1,2,4,3 or the

reverse of one of these. Now choose two leaves p and p* which are in the same fiber H, after

the singularity on T. Since fl and /?* are disjoint arcs in H,, the endpoints 2 u 3 of /? cannot

separate the endpoints 1 u 4 of /?* on A = dH,. But then the order must be 1, 2,3,4 or its


1 4

Fig. I1

reverse, as claimed.

Our next general observation is that if V, E and F denote the number of vertices, edges

and faces in the tiling, then V = F and E = 2F. This follows from the fact that each tile has

four edges and each edge is adjacent to exactly two tiles, so E = 2F. Then the Euler

characteristic of T is 0 = V- E + F = V - F, and the assertion is proved.

TO continue, we assume that T has been cut open to a tiled fundamental domain R

which is a rectangle of dimensions k tiles by k’ tiles. If T is a tile in R, then if we cross any

edge of T we will encounter a tile of opposite sign, so we may assume without loss of

generality that R has been chosen so that k 2 2 and k’ 2 2. We now consider Theorem 1 in

the special case when k = k’ = 2, so V = F = 4. All four tiles have the same four vertices,

and from this it follows (see the left picture in Fig. 12) that the cyclic order of the vertices

around each is alternately 1,2,3,4 and 4,3,2,1. Notice also (the left picture in Fig. 12 again)

that the sense of increasing 0 is clockwise about 1 and 3 and counterclockwise about

2 and 4.

Choose four points on the oriented axis A and declare them to be 1,2,3,4 in that order.

Next, choose little disc neighborhoods of the vertices, and attach them to the axis. If the two

sides of the oriented surface T are painted red and green, the fact that the sense of the flow is

alternately clockwise and counterclockwise about 1,2,3,4 shows that the axis A will pierce

T from the red, green, red and green sides at 1, 2, 3, 4.

The next step is to embed the singular leaves in 3-space. By hypothesis, the four

singularities occur at four distinct values of 8. Choose four distinct fibers of H, say at 8,) 8,,

6, and 8, in that cyclic order in the fibration, and put down the pair of singular leaves

(joining 1 to 3 and 2 to 4) in each fiber. Next add an arrow at each of the four singular points

pointing in the direction of increasing or decreasing 8, according as the singularity is

positive or negative. See the right picture in Fig. 12. The front left and rear right tiles have

positive signs and the front right and rear left ones have negative signs. The remaining

regions of T are everywhere transverse to the leaves of the foliation and there is a unique

way to attach them. This proves Lemma 8, in the case where k = k’ = 2. Notice that in this

case T is a union of two “tubes”, each of which is made up out of two bb-tiles.

Our next claim is that in the general case either k or k’ must be 2. We have already

shown that we may assume that k, k’ 2 2, so to prove this assertion we assume that k’ > 2

and k > 2 and arrive at a contradiction. Choose a non-singular fiber H,. A b-arc /3 in H, will

be called innermost if one of the sides of the axis A split along afl contains no other points of

A n T. Choose an innermost b-arc /I in H,. Without loss of generality we may assume that p

is a common edge of a pair of adjacent tiles Tl and T, in the tiling of T, as in Fig. 13. The six

vertices of T, u T2 are necessarily distinct because k’ > 2, k > 2. Label them 1,2,3,4 reading

clockwise about the boundary of, say, T2 and 6, 5, 4, 3 reading clockwise about the

boundary of T,. We wish to determine the order of these six vertices on A. As noted earlier,

the four vertices 1, 2,3,4, and also the four vertices 3,4, 5,6 must have that cyclic order on

A, up to a possible reversal of orientation. Assume one of the two possible orientations on A


Fig. 12.

6 3 2

The tiles TI and T,

4

3

0

5

6

Fig. 13

and assume that the locations of 3, 4, 5, 6 on A have been fixed, as in the right picture in

Fig. 13, which shows the axis A as the boundary of the disc fiber H, at 9 = 3.

We ask where 1 and 2 can go? Going to the top left sketch in Fig. 13 we notice that there

must be arcs b,,(8), bJ4(0) and b12(0) joining the vertices 5 to 6, 3 to 4, and 1 to 2

respectively in T n H, at 8 = 3. Since H, is a disc, this means that 1 and 2 cannot separate

either 5 and 6 or 3 and 4 on A = dH,. Going to the top right sketch in Fig. 13, let Aij denote

the subarc of the axis which lies between adjacent vertices i and j and does not contain the

other two pierce points. The fact that p is innermost also shows that 1 and 2 cannot be in

A,,, so either 1 is in A,, and 2 is in A,, or both are in A,,, A,, or A,,. The second and the

fourth cases are equivalent under a symmetry, so it suffices to consider the first, second and

TOP33:3-J


third. However, by reversing the order of the fibration, we interchange the roles of 1,2 and 5,

6, so the first case is equivalent to the third. Thus we may assume that the points 1,2 are in

A,, or A,,. The six vertices thus have the order 1,2,3,4,5,6 or the order 3,4,5, 1,2,6, as

illustrated in the two bottom sketches in Fig. 13.

Passing to Fig. 14, we consider the case where the order on A is 1, 2,3,4,5,6. We now

look at the two tiles which are glued to TI and T, along the 632 edge of Tl u Tz. Let the new

vertices be 5’, 4, l’, so that the new tiles are 65’4’3 and 34’1’2, as in Fig. 14. The facts that

there are no tile vertices between 3 and 4 on the finite part of A, and that the signs of the four

tiles alternate checkerboard fashion, and that k’ > 2, k > 2 forces the 9 vertices to be distinct

and to have the order l’, 1, 2, 3,4, 4, S, 5,6 on A, as illustrated in Fig. 14. (The tiles 65’4’3

and 34’1’2 are “behind” 3456 and 4321).

Since k’ > 2 and k > 2 there must be at least one more new tile glued to Tl along its 54

edge. Let its new vertices be X and Y, so that the new tile is 54YX. The fact that T is

embedded forces X to be between 6 and 5’ and Y to be between 4’ and 3. The fact that the

signs of the tiles 3456 and 54YX are different forces X to be between 6 and 5 and Y to be

between 4 and 3. However, /I = 34 is innermost, so that is impossible. Thus the only

possibility is that 5 = 5’, 4 = 4, 1 = 1’. The case where the order of the six vertices is 3,4,5,

1,2,6 is identical, even to the choice of the additional tiles which are glued on. Thus we have

proved that R has dimension 2 tiles by k tiles.

How do our 2k tiles fit together? The fact that the signs of the singularities alternate in

the tiling of T will be seen to place yet another restriction on the embedding, i.e. we will see

that “tubes cannot go through tubes”. (Remark: without this restriction it is possible to

construct examples which contain tubes going through tubes, however any such example

will necessarily contain adjacent tiles with singularities having the same sign.) To see this,

we pass to a slightly different decomposition of T. Go back to the fundamental domain R, which is a rectangle which is 2 tiles high and k tiles wide. Choose vertical arcs c 1 , . . . ,ck on R which run from the top edge of R to the corresponding point on the bottom edge, each ci

containing two singular points of the foliation. Thus each ci is a circle on T. Let A,,. . .,A,

be annular neighborhoods of cl,. . .,ck on T. Let S,,. .,S, be the complementary annuli.

See Fig. 15. Then each Si is a 2-holed sphere which is pierced twice by A and is foliated

without singularities. We may visualize the Ats as “joining tubes” which join up the Sj’s in a

cycle.

We now concentrate on how a single Si and its two joining tubes Ai and Ai+I are

embedded in S3. We have labeled various leaves in the foliation in Fig. 15 as occurring in

6 2

Fig. 14.


Fig. 15

fibers at 0 = 1, 2, 3, . . . ,12, also we have labeled the signs of the singularities. Figure 16

shows these same leaves as they would appear on Ai u Si u Ai+ 1 in 3-space. A key point in

determining the embedding is the checkerboard arrangement of these signs.

We now have a standard “template” from which we can construct the entire embedding

of T. The embedding is determined by the embedding of our standard template. Continuing

the construction of T in this manner we see that each Ai can be thought of as the boundary

of a regular neighborhood of an arc ai contained in a disc fiber HOi. Furthermore, the

manner in which copies of the template are glued together (see the example in Fig. 17)

guarantees that these arcs are pairwise disjoint. That is, T has a type k embedding. I(

93. TORI WHICH ADMIT MIXED FOLIATIONS

In this section we concern ourselves with mixed foliations. Recall that a mixed foliation

will have both type bb and type bc singularities. We say that a mixed foliation is standard if

all of its singularities are type bc.

LEMMA 9. Assume that T is a torus in the complement qf a closed braid, and that T has a mixed foliation. Then, after a sequence of exchanges we may assume that T has a standard

mixed foliation.

Proof: The proof will be similar to the proof of Lemma 7, but it will be more

complicated. Since a mixed foliation does not naturally give rise to a cellular decomposition

of T, we must construct one.

The first thing to notice is that bc-annuli must occur in pairs, because the c-circle

boundary component of a bc-annulus can only be attached to c-circle boundary component

of another bc-annulus. Now, consider the annulus, W, which is constructed by attaching

two bc-annuli together along their c-circle boundary component, as in the top picture in

Fig. 18. Each component of a Wis the union of two b-arcs and contains two points of A n T. We now cut W open along two new disjoint edges, e and e’, each having its endpoints at

points of A n T which are on distinct components of W, as in the bottom picture in Fig. 18.

We see that W is the union of two be-tiles where the boundary of be-tile has two b-arcs and

two of the new e-arcs. Notice that the number of be-tiles constructed in a mixed foliation is

exactly equal to the number of bc-annuli. The vertices in the be-tiling are the points of

A n T. We now have a cellular decomposition on T where the O-cells are the points of A A T,

the l-cells are b-arcs and e-arcs, and the 2-cells are bb-tiles and be-tiles. Let V be the number


Fig. 16.

7(>

80 ; I

6 3

LI

7 2

8 5 4 1

Fig. 17

of O-cells, E the number of l-cells, and R the number of 2-cells. Then V - E + R = 0. As

before, since each 2-cell has four l-cells and every l-cell is adjacent to two 2-cells we know

2R = E. As before, V and E are related by 2 V - E = 0. We could easily obtain an analog of

equation (l), but we need something more, because we have two kinds of edges, b-edges and


P' P .

Fig. 18.

e-edges, and need to distinguish between them in our count. Let I’(/?, E) be the number of O-

cells in the tiling of T which have fi b-arcs and E e-arcs incident to them. Let i be the valence

of a O-cell. Then fi + E = i, so that V(/I, i - fi) denotes the number of O-cells of T which have

valence i and have /I b-arcs as l-cells. Notice that i - /I is at most p in the expression

I’(/?, i - fi), because b-arcs and e-arcs alternate around the boundary of a be-tile, and a bb-

tile contains only b-arcs in its boundary. Thus, i 5 2p I 2i or [i/2] I j3 I i. Thus V(i) is the

sum of all P’(fi, i - j?), as fl varies between [i/2] and i. Therefore the analog of equation (1) is:

2V(2,0) + 2V(l, 1) + V(2,l) + V(3,O) = f f: (i - 4)V(/?, i - /I) i=S p=[i/Z]

where as before both the left hand side and the right hand side are non-negative.

We now claim that V( 1,l) = 0. For, suppose not. Then there is a vertex which is incident

to a single b-arc and a single e-arc. This can happen only if there is a bc-annulus J which has

both of its b-arc boundary edges identified in the foliation. Let c be the c-circle boundary

component of J which is contained in a fiber H, of H. Let d, be the subdisc in H, which c

bounds. Notice that d, is punctured by L, and the number of such punctures is algebraically

non-zero. Construct the 2-sphere J u d,. This sphere has non-zero algebraic intersection

number with L, but that is impossible.

We also claim that V(2,l) = 0. If not, there is a vertex p of type (bbe), i.e. the three tiles

which meet at p have singularities of types b, b and e. Let b, b’ and e be the tile edges of type

b, b, e which meet at p. Then e necessarily cuts between the two boundary components of an

annulus W as in the bottom picture of Fig. 18, so b and 6’ must be in the situation of the

edges b and b’ in the bottom picture of Fig. 18. Also, since there are exactly three tiles which

meet at p, and since two of them are type be, the third tile must be a bb-tile. However, a bb-

tile has four distinct vertices, whereas b and b’ have two endpoints in common,,namely p and

also p”. This means that the bb-tile would have three vertices, not four, so this cannot occur.

Suppose now that there are vertices having valence 2 5. Then the right hand side of

equation (2) is positive, so the left hand side must be too. Since we have just proved that


V(1, 1) = V(2,l) = 0, we conclude that either V(2,O) > 0 or V(3,O) > 0, i.e. there is a vertex

of type (bb) or (bbb) in the mixed foliation of T. Part (i) of Lemma 6 applies. Thus L admits

an exchange move, and after the exchange move we may reduce the complexity. This

process ends when every vertex has valence 4.

By hypothesis, T has a mixed foliation, so e-edges occur. If every vertex has type (bebe)

the mixed foliation is standard and we are done. If not, some bb tile must be adjacent to a be

tile, so we may assume that there is a vertex of type (bebb). The vertex p in Fig. 19 illustrates

the only possible picture. There are two bb tiles and two be tiles. If we now translate this

configuration back to the foliation on T we see that p will have two bb tiles and one bc

annulus attached to it, as shown in Fig. 19. By an analysis which is similar to that used in

the proof of Lemma 7, we deduce that the signs of the two bb tiles are opposite. Thus the

sign of the singularity in the bc annulus will agree with that in one of the bb tiles, as

illustrated. Lemma 6 applies again. The foliated region R which we wish to change now is a

union of a bb-tile and a bc-annulus. See Fig. 20. In the original foliation, depicted on the left,

the first singularity is between b-arcs joining p1 to p4 and p2 to p3. After the singularity there

will be a b-arc joining p1 to pz, which forms a singularity with the c-circle. After the change,

the first singularity is between b-arcs joining p2 to p3 and the c-circle. This results in a new

b-arc through p2 and p3, which then has a singularity with b-arcs which join p1 to p4.

After the change in foliation the vertex p1 will have valence 2. By Lemma 6, part (ii) we

may then reduce the complexity. After a finite number of such reductions we will obtain a

torus which has a standard mixed foliation. II

p has typdbebb)

Fig. 19.

isotopy ---=?

(a)

Fig. 20

(b)


To continue, we assume that we are given a collection {T, ,. . .,Tk} of disjoint tori, each

of which has a standard mixed foliation. Choose any torus T in the collection. Cutting T

open along its c-circle boundary curves, we obtain a collection of k annular regions, each

punctured twice by the braid axis, as in Fig. 21, and each containing two singularities of the

foliation. Divide each such region into three concentric annuli, two of which (call them Xi

and Yi) are closest to the boundary components xi and yi, and are foliated without

singularities by circles parallel to xi and yi. We choose to think of the central annulus Si as a

sphere with holes. It is foliated with two singularities at points we have labeled pi and qi.

Choose the notation so that Yi and Xi+ 1 are adjacent on T, and let Ai be the annulus

YiUXi.,. Then T is the union of the 2-spheres S, , . . ,Sk, each punctured twice by A and

containing two bc singularities, tubed together along the tubes A,,. . .,A,. The cyclic order

of the spheres and tubes on T is . . .Si_ 1, Ai, Si, Ai+ 1 ,. . . . The bottom picture in Fig. 21

shows the singular leaves in the foliation of Si.

To understand the embeddings in 3-space, we pass to Fig. 22. The embedding of the

tubes is clear, because each is foliated without singularities by c-circles. Thus they are

embedded as braided tubes (see Fig. 22). Each tube may go around the axis several times,

also distinct tubes may be braided with one-another.

As for the spheres, regard the braid axis A as a line in 3-space and fibers of H as half-

planes through A. Choose two points on A and declare them to be the two points 1 and 2 of

A n Si. Choose distinct half-planes at, say, 8_ 1 and el, and declare them to be the two fibers

(a) Sphere-tube decomposition

(b) Foliation of sphere-part

Fig. 21


embedding of tube part of T in 3-space

3 to si+j

embedding of sphere

part of T in 3-space

Fig. 22

which are tangent to Si. Join the punctures by singular leaves in these two fibers. The

remaining leaves of the foliation of Si will non-singular, and will be either be b-arcs joining 1

and 2 or circles. By exactly the procedure that we used in the proof of Lemma 8 we fill these

in. There is essentially one way to do it, as Si is transverse to the fibers of H everywhere

except at 8, and I!_ 1. See the bottom pictures in Fig. 22. Figure 23 illustrates an example of

a torus T which has tubes passing through tubes. An associated H, sequence is illustrated in

Fig. 24.

LEMMA 10. For any tube Ai, the singularities on the 2-spheres Si_l and Si which are

closest to Ai have opposite signs, as in Fig. 25.

Proof: We examine the situation in a sequence of fibers, starting just before the

singularity in Si, until just after the singularity in Si+ 1. We use the pictures in the top row in

Fig. 26 to illustrate what we wish to say. The positive bc-singularity at & 1 results in the

creation of the tube Ai. The circle which is illustrated at 8, is the intersection of Ai with the

fiber at BO, and the fact that the singularity which created it was positive implies that the side

of it which does not bound a disc is negative. But then the bc-singularity at 8,, which

annihilates the tube, must be negative. II

A tube Ai in the decomposition of T will be said to be outermost if, for any non-singular

fiber H, which meets Ai non-trivially the c-circle H, n Ai is not contained in the disc

bounded by H, n A, for any other tube A,.


Fig. 23

start

Fig. 24.

LEMMA 11. (sphere amalgamation lemma). Choose an outermost tube Ai in the de-

composition of T. Suppose that there exists a rectangular region R in all of the disc fibers

between OK, and 6,, as illustrated in the middle picture in Fig. 26, such that for all

0 E CO_ 1, O,] the link L intersects H, n R only inside the subdisc bounded by the c-circle which

is inside R. Then after a change infibration we may replace the be-singularities in the3bers at

Otl with bb-singularities. Moreover, after the change we will have a new standard mixed

foliation with fewer bc-singularities.


Fig. 25.

Fig. 26.

Proof Refer to the middle diagram in Fig. 26, where the rectangular region R is

illustrated. Its boundary is labeled c[ u /I u y u 6. The hypotheses of the sphere amalgam-

ation lemma ensure that there is no obstruction to replacing the HO-sequence in the top row

of Fig. 26 with the sequence in the bottom row. A careful look at that sequence shows that

after the change in fibration the vertices which are labeled 1, 2, 3, 4 will all have valence 2

and type bb. We may then use Lemma 6 to eliminate these bb-singularities, to produce a

standard mixed foliation with fewer he-singularities. II

LEMMA 12. (the exchange move lemma for tori with standard mixed foliations). Let Ai

be an outermost tube and let Si_l and Si be its adjacent 2-spheres. Then, after a series of


exchange moves, we may assume that Si_ 1, Ai, Si satisfy the hypotheses of the sphere

amalgamation lemma.

Proof: We examine the geometry of Fig. 26, but from a new point of view: instead of

looking at how L, Si_ 1, and Si meet a sequence of fibers of H, we look at them under a

regular projection onto a plane which is orthogonal to A. See Fig. 27. Regard S3 as

R3 u ( CC } and A as the z-axis u { cc }. Let cli be the core of the tube Ai, i.e. cli is an arc in S3

- L which joins the capped spheres CSi_ I and CS,. Without loss of generality we may

assume that the orthogonal projection rt of R3 onto the x-y plane induces a regular

projection for the link L and the arc c(~. We may further assume that the images Di_ i =

x(CS,_,) and Di = rr(C&) are radially foliated concentric discs and that neither L nor mi

passes under or over Di or Di_ I. Strands of L will, however, enter the discs, and then wind

about the axis, possibly braiding as they do so. There may be other tubes Aj in the picture,

but if so we may assume without loss of generality that they are arbitrarily close to the

strands of L, because our tori are incompressible, so for simplicity we have not shown them.

There may also be strands of L inside the tube Ai, but if so they may be assumed to be

arbitrarily close to cli, so for simplicity we have not shown them either in Fig. 27. Finally, we

will simplify our picture whenever we need to do so by pushing the braiding of L into

“boxes”, as needed, so that the visible part of L is seen as a set of parallel strands in our

projection.

We now observe that if the arc ai lies above all of the strands of L in Fig. 27, then we may

push it into upper half-space. The intersections of a regular neighborhood of cli in 3-space

with fibers of H will then give us the region R which was described in the sphere

amalgamation lemma. In general that will not be the case. Instead, xi will thread under and

over L in its passage from Di_l to Di. Our problem is to eliminate that threading.

In order to get a measure of the complexity of the threading of our arc cli, we now lift

Di_ 1 and Di a little bit out of the plane of projection, so that they are seen as disjoint discs

Fig. 27.


above and parallel to the projection plane. This “lifting” of Di_ 1 and Di will lift the

endpoints of ai out of the projection plane, so that now G+ starts at dD,, above the plane of

projection, and then (as it crosses under L) it will be forced to dip down, puncturing the

projection plane. We assume that it dips up and down, puncturing the plane transversally

each time, until it ends at aIIi. Without loss of generality we may assume that the punctures

occur away from the braid boxes. See Fig. 28. Define the complexity C(q) to be the number

of points of intersection of cli with the x-y plane. By our earlier observation, if we can find

moves which reduce C(q) the lemma will be proved.

Our first move is a rotation of Di_ 1, as illustrated in Fig. 29. We have not shown Di in

this picture, however we have added the braid box which is inside Di_ 1 and the correspond-

ing strands of L. The purpose of the rotation (look ahead to Fig. 30) is to create an interval

in the polar angle function belonging to the braid box which is inside Di_ 1 which is disjoint

from the corresponding braid box inside Di. As a result of the rotation the portion of L

which is out of the plane of projection and attached to aII_ 1 will be lengthened, while the

corresponding portion of CQ (and all parts of L and other tubes which are close to cq) will be

shortened. Rotating Di_l does not change C(CQ).

Our next move is an exchange move, illustrated in Fig. 30. The capped 2-sphere which we

are calling Di_ 1 is pulled across the axis, taking the part of the link which is inside it across

the axis as it goes, and pulled through L, and then back across the axis. The orders of Di_ I

(which is illustrated) and Di (which is not) will be switched. This will in general introduce

new braiding in L, as illustrated. However, the important feature is that after the exchange

move we will be able to isotope zi so that it has one less puncture with the x-y plane.

After a finite number of rotations and exchange moves we will have C(q) = 0, and

Si _ 1 u Ai u Si will satisfy the hypotheses of the sp’rere amalgamation lemma. II

After applying Lemmas 11 and 12 as often as necessary, we will have replaced our torus

T by a new embedded torus which has a standard mixed foliation and exactly one 2-sphere

and one tube, that is it has a type 1 embedding. II

$4. THE PROOF OF THEOREM 1, COROLLARIES 1 AND 2 AND THEOREM 2

Proof of Theorem 1. By hypothesis, we are presented with a finite collection 51

= {T,,. . .,T4} of tori in S3 - L, where the ith torus Ti has a foliation whose type can be

altered to type .si = 0, 1 or k. The key to the proof is to do the modifications which we

introduced earlier in a controlled order.

By Lemma 4 we may assume that every torus has a circular foliation, a mixed foliation

or a tiling. Also, we may assume that if the foliation is circular then the embedding is type 0.

The modifications which were used to alter the tori Ti with si = 1 and k 2 2 were

described in Lemmas 6, 7, 8, 9, 11 and 12. They used four different modifications:

Modijcation (i): Isotopy of 3-space;

Modijication (ii): Change in the fibration H, predicated on the existence of two adjacent

bb-tiles having the same parity;

Modification (iii): An exchange move and elimination of a valence 2 vertex, if such a

vertex exists;

Modijication (iv): An exchange move and rotation, as in Lemma 12.

Modification (i) does not change the embedding types of any of our tori. Modification (ii)

alters the foliation of two bb tiles, but leaves everything fixed outside an arbitrarily small


projection plane

Fig. 2%.

projection pIalE \

Fig. 29

neighborhood of these tiles. Modification (iii) alters the foliation in a neighborhood of the

valence 2 vertex which is being eliminated, but leaves everything outside a small neighbor-

hood of this vertex unaltered. Modification (iv) changes the embedding of the tori of type 1

and k, but not of a torus of type 0. We used modifications (i), (ii) and (iii) in standardizing the

embedding of the tori of type k. We used all four modifications in standardizing the

embedding of the tori of type 1.


exchange move

Fig. 30.

As noted earlier, to begin we assume that every torus has a type 0 embedding or has a

mixed foliation or a tiling. Applying the four modifications as often as necessary (using

Lemmas 9, 10, 11, 12) we may assume that every 2’; with si = 1 has an embedding of type 1. Notice that if there were tori which admit tilings, then after the modifications they will

continue to admit tilings, however the tilings themselves may change. Using Lemmas 4, 6

and 7 we then go to the tori Ti with si = k, and standardize their tilings. By Lemma 8 we

conclude that every torus of type Ei = k will then have a type k embedding. The modifica-

tions used in Lemmas 4,6 and 7 are (i)-(iii) of the above list. These modifications only alter

the foliation of the tori locally, i.e. within two bb-tiles for operation (i) and within a

neighborhood of a valence 2 vertex for operation (ii). Therefore they will not disturb the

embeddings of type 0 and 1. The proof of Theorem 1 is complete. II

In the Introduction we stated two results (Corollaries 1 and 2) which give detailed

information about the closed braid representatives of non-simple links of braid index 3

and 4. We are now ready to prove them.

SPECIAL POSITIONS FOR ESSENTIAL TORI IN LINK COMPONENTS 5.53

Proof of Corollary 1. We are given a non-simple link L of braid index 3. Choose a closed

3-braid structure for L. Let T be the essential, non-peripheral torus in S3 - L. By Theorem

1 we may assume that T has an embedding of type 0, 1 or k relative to the half-planes {H,;

19 E [0,2x]} in our braid structure. Choose a non-singular fiber H, of H. Our link L will

intersect H, in 3 points, which we refer to as “dots”. We examine T n HB when T has a type

0,l or k embedding. The reader may find it helpful to consult Fig. 6, which depicted T n H,,

as viewed on H, in the three cases.

Our first claim is that T cannot be type k with k 2 2. For, suppose that it was. The

intersections of T with H, will consist entirely of b-arcs and there are k I 2 of them, with 2k

endpoints on A = aH,. We study them, as they evolve in the cycle of fibers {H,; 0 < 0 I 27~).

Each b-arc must eventually be surgered, and during the cycle there are exactly four singular

arcs which emanate from each point of A n T.

Assume first that k = 2. There are 4 points in A n T and there are two b-arcs in each

non-singular fiber. Choosing a non-singular fiber H,, we see two b-arcs with their endpoints

on the axis A. Let us suppose that the notation is chosen so that the points in A n T are 1,2,

3,4 and that b,,(B) and b34(Q) are the b-arcs in that fiber, where bij joins i to j. They divide

the fiber into two half discs di2(8) and d3.+(0) and a half annulus A(8) and there must be a

dot in each half-disc because if not our b-arcs would be inessential. That means that there is

at most one dot in A(8). But then, after the first surgery we will have new b-arcs br4(@) and

bz3(O’), which divide H, into two half-discs and a half-annulus, and on of the discs will not

contain a dot. Thus there will be an inessential b-arc. Hence type k is impossible for k = 2,

and therefore even more so for k > 2.

We rule out the occurrence of a mixed foliation on slightly different grounds. If there is a

bc-annulus in the foliation of T, then the c-circle associated with this annulus must be

essential. It therefore bounds a subdisc d, of H, which contains one of the three points of

H, n L. If we surger T along d,, we produce an essential twice punctured 2-sphere. But then,

L must represent a composite link, contradicting the assumption that L is prime. Therefore

T must have a standard circular foliation. The description of L follows immediately, because

there are only three braid strands. II

Proof of Corollary 2. We proceed as in the proof of Corollary 1, only now our link L will

intersect each H, in 4 points. We examine T n H, when T has a type 0, 1 or k embedding.

Type 0. Notice that T n H, contains at most two c-circles. Each bounds a disc which

contains at least one dot. Thus there are three cases:

Case 1. There are two discs, each containing two dots. This gives the 4-braid

(a,)P(a,)*(a,a,a,a,)‘, where the factors (al)” and (G~)~ correspond to braiding inside the

discs and the factor (aza, c3 cJ, corresponds to a knotting of the torus by braiding between

the two discs. Case 2. There is one disc d. It contains three dots. The fourth dot is in H, - d. This gives the 4-braid p(~~g~~:cr~rr~)~. The braid /? corresponds to braiding among the

three dots inside d. The factor (03g2~:cr2~3)p represents a winding of the fourth strand

around the outside of the unknotted solid torus. Case 3. There is one disc d which contains

two dots. The other two dots are in H,\d. The braid is (r~,)~(o,~~a,)~(o,a~o,)‘. The factor

(cr2)P comes from braiding inside the disc. The factors (~~rri~i)~ and (a,o<a,)’ represent

separate windings of the remaining two strands about the unknotted torus T.

Type 1. The intersections of T with H consist of a single b-arc and some number of c-

circles. The arc cuts off a half-disc h in H, which contains at least one dot. The circles each

bound discs, and each disc contains at least one dot, also the same number of dots are in

each disc. There also may be dots outside h and all the discs. The total number of dots is 4.


This shows that there cannot be more than one disc, for if there were two, there could only

be one braid strand in each, but then T would be peripheral. This shows that T is unknotted,

so there must be a dot outside h and d, leaving at most two dots inside d. However d must

contain at least 2 dots, for if it only contained 1 then T would be peripheral. Thus the only

possibility is the braid b(ol ~:a,)~, where the fourth strand winds about the axis inside the

sphere, the middle two are inside the tube, and the factor (a, a<~,)~ corresponds to the first

strand wrapping about the middle two, to make T incompressible.

Type k. The intersections of T with H, consist entirely of b-arcs and there are k 2 2 of

them, with 2k endpoints on A = aH,. We study them, as they occur in the cycle of fibers

(H,; 0 I 8 I 2n). As in the case of 3-braids each b-arc must eventually be surgered, and

during the cycle there are exactly four singular arcs which emerge from each point of A n T.

Assume first that k = 2. We will show that we obtain the example depicted earlier in

Fig. 3. Since k = 2, there are 4 points in A n T and there are two b-arcs in each non-singular

fiber. Choosing a non-singular fiber at B,, let us suppose that the notation is chosen so that

the points in A n T are 1, 2, 3, 4 and that b,,(8,) and b,,(O,) are the b-arcs in that fiber,

where bij joins i to j. They divide the fiber into two half discs d,,(O,) and d,,(O,) and a half

annulus A(8,) and there must be a dot in each half-disc because if not our b-arcs would be

inessential. There must also be two dots in A(8,) for if not then after the first surgery we

would obtain an inessential b-arc. Braiding can occur between the two dots in A(0,)

between adjacent surgeries. After the first surgery we will have new b-arcs bi4(0) and bz3(0),

which again divide H, into two half-discs, each containing a dot, and a half-annulus

containing two dots, and again there can be braiding. Continuing in this way we obtain

blocks of elementary braids (a,,)P(a,,)r(o,,)4(a~~)s. . However, there must be exactly four

syllables because each block of braidings occurs during the &values between two adjacent

singularities, and there are exactly four singular leaves emerging from each of the points 1,2,

3, 4 of An T. Thus the only possibility is the 4-braid (a,,)p(a,,)r(o,,)4(a,,)s.

The case k > 2 is ruled out for the same reasons as were used to rule out the case k = 2

for 3-braids, i.e. it is impossible because there will necessarily be an inessential b-arc. (/

In Theorem 1 we studied special positions for the torus T in S3 - L, relative to the braid

axis A and the fibers of H. The fact that T is an embedded torus in S3 implies that T bounds

a solid torus V on at least one side. Theorem 2 asserts that, in each of the three types of

embeddings, there is a natural way to describe the inclusions of L n V and A n V in V. See

the Introduction for the precise statement.

Proof of Theorem 2. In the case of type 0 embeddings each c-circle c(Q) lies in both T and

some fiber H,, and so bounds a disc d,(B) in H,. The number p of such discs in H, is

independent of 8, as is the number 4 of points of intersection of K with a single disc. The

union of the d,(B)‘s thus spins out a p-strand braided solid torus V, p 2 1, as 8 is varied

between 0 and 271, and K is a closed w,-braid inside V. The axis A does not meet V. The solid

torus V is as depicted in Fig. 5.

In the situation of type 1 embeddings we study a sequence of fibers H, of H to

understand how L n V is situated in V. See Fig. 31. The first picture shows a typical non-

singular fiber H, of H. Its intersections with T will be one b-arc and some number of c-

circles ci,. . .,cp, the number corresponding to the number of times the single tube wraps

around the axis A. Each non-singular c-circle c bounds a disc d,(e) in H, and the b-arc

bounds a half-disc hb(0). The links meets all of these, for if not there would be inessential c-

circles and an inessential b-arc. The solid torus V is swept out by the hb(Q) and

d,, (e), . . . dJ0) as 8 varies over the interval [0,271]. Notice that as we push H, forward in the


@+@ff+,fg) Cycle of fibers for type I embeddlng

Fig. 31.

Part of HO sequence for type k embedding

Fig. 32.

fibration the discs d,(B) may move around each other, creating a braided tube. Thus V is a

3-ball with a (possibly braided) handle attached to it. The axis A meets the 3-ball part of V in

a single arc.

The b-arc and c-circles separate the points of L’ n H, into sets such that as 6 increases

braiding only occurs between points of L’ n H, which are in the same set. The only values of

8 at which points in different sets are allowed to braid is between the two occurrences of bc

singularities. It is clear that all of the braiding in this part of the cycle can be pushed into a

single block. Thus we obtain the picture given in Fig. 5.

If it should happen that V’ = S3 - V is unknotted, the identical argument gives a

symmetric picture for the inclusion (L u A) n V’ in V’ in the case of a type 1 embedded torus.

To understand L in the situation of type k embeddings, k 2 2, we look at that part of the

H, sequence which is associated to the foliation of a single annulus Ai and parts of its

adjacent 2-spheres. Figure 32 shows two b-arcs, and we follow them in a sequence of fibers

of H which includes a “ + ” singularity followed by a “ - ” singularity. At first we will see

two subsets of L n H,, where each is contained in a region which one of the b-arc splits off in

TOP33:3-K


H,. Braiding can occur, but only between the arcs in each set. After the “ + ” singularity the

two subsets of L n H, will merge into a single region of H, - (T n H,). It will then be

possible to have braiding between the two sets. After the subsequent “ - ” singularity the

points of L n H, contained in this single region will break up into subsets again.

The solid torus V is a union of the 3-balls swept out by the regions bounded by the

b-arcs, joined up by tubes, where the core of each tube lies in a single fiber. There will be k

3-balls, each meeting A in a single arc. There will be a block of braids associated to each

3-ball, and another block associated to the connecting annulli, but we may amalgamate

them as in Fig. 5. Our proof is complete. II

Acknowledgements -The applications of our work to companionship were suggested by a question asked by the

referee, whom we thank for stimulating us to think about this special case. We also thank Daniela Renate Hass,

who as an MA student at SUNY Buffalo had the geometric insight to realize that a single 2-sphere and annulus

suffice in the case of type 1 embeddings. Lemmas 10, 11 and 12 are based upon that insight.

REFERENCES

1. D. BENNEQUIN: Entrelacements et equations de Pfaff, Asterisgue 107-108 (1983), 87-161. 2. J: BIRMAN and W. MENASCO: btudying links via closed braids I.

(1990), 115-139. .~ I@plit and composite links Invent. Math. 102

f

3. W. H. JACO and P. B. SHALEN: Seifert fibered spaces in 3-manifolds, Memoirs AMS, 21, (1979) Num. 220.

4. K. JOHANSSON: Equivalences d’homotopie des vari&ts de dimension 3, C.R. Acad. SC. Paris, 66 (1975),

1009%1010.

5. D. ROLFSEN: Knots and Links, Publish or Perish, (1974).

6. H. SCHUBERT: Knoten und Vollringe, Acta Math. 90 (1953), 131-226.

7. W. THURSTON: On the geometry and topology of 3-manifolds, Lecture notes, Princeton Univ., (1982),

unpublished.

8. S. YAMADA: The minimum number of Seifert circles equals the braid index of a link, Inu. Math. 80 (1987),

347-356.

Department of Mathematics

Columbia University

New York, N.Y. 10027

U.S.A.

Department of Mathematics

University at BufSalo

BufSalo, N.Y. 14222,

U.S.A.

Added in proof.

After this manuscript went to press our tireless referee requested additional clarification of the formula

b(L) = w1 + ... + wx of Corollary 3. The point which troubled the referee was that the integers w,, . ,w~, and thus the braid index b(L) of the non-simple knot L, did not appear to depend upon the manner in which the solid torus

V, is embedded in 3-space. However, that is a misinterpretation of w,, .wk. The example which is given in Fig. 7 and 8 may help in understanding the situation. The knot L in Fig. 7 is a

double of the trefoil C, which is depicted in Fig. 8 as a union of 5 planar arcs, each of which begins and ends on A. Notice that the embedding of C which is given in Fig. 8 was chosen so that its double L would be a closed braid

with respect to A. However that is not all; in addition, L must lie inside a solid torus neighborhood V, of C and the solid torus V, must be positioned (using the main results in this paper) to exhibit the fact that its boundary T, is a

type 5 torus.

This example illustrates the general situation. The numbers w 1,. ,wk depend upon inter-related phenomena, i.e. the way in which L is embedded in V,, and the way in which A meets Vc The facts that (i) L is a closed braid with respect to A, and (ii) L lies inside a solid torus neighborhood V, of its companion C, and (iii) aV, has a type k embedding, combine to place inter-related constraints on the integers wI, ,wk. Thus b(L) does, in fact, depend

upon both the embedding of L in V, and the embedding of C in 3-space.

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SPECIAL POSITIONS FOR ESSENTIAL TORI IN LINK …jb/essential-tori.pdf · SPECIAL POSITIONS FOR ESSENTIAL TORI IN LINK COMPONENTS 527 Example of type 1 embedding Fig. 2. Example of